Is the Boltzmann Equation in GR Truly Coordinate Independent?

[DrFaustus]

Hi everyone,

have a question about covariant, coordinate independent quantities in GR.

Reading Kolb and Turner's book The Early Universe one can find the Boltzmann equation in a GR setting. Now, one of the terms in that equation is
$$- \Gamma_{\alpha \beta}^\gamma p^\alpha p^\beta \frac{\partial f}{\partial p^\gamma}$$
where $$\Gamma$$ is the Christoffel symbol, $$f = f(x^\mu, p_\mu)$$ is the particle distribution function, which is a scalar and depends on the coordinates $$x^\mu$$ and the momenta $$p_\mu$$.

My question now is simply how to see that this is a coordinate independent quantity? The Christoffel symbols do not transform like tensors, but the above formula really should be a scalar quantity, like the index notation suggests. So my guess is that the partial derivative with respect to the momenta must transform in a way so to make the above a scalar. Is this correct? Can you explain in some detail? Or give me some references to check it out?

Thanks!

 

[bcrowell]

Even ignoring the fact that the Christoffel symbol isn't a tensor, there's also the issue that the derivative is not a covariant derivative. It might be easier to figure this out if you give us the whole equation.

 

[DrFaustus]

It's the collisionless Boltzmann equation:

$$p^\mu \frac{\partial f}{\partial x^\mu} - \Gamma_{\alpha \beta}^\gamma p^\alpha p^\beta \frac{\partial f}{\partial p^\gamma} = 0$$

The first term is manifestly covariant, and so should be the second one.

As for the Christoffel symbol, Wald states it is a tensor. The reason why it does not transform like one might suspect is that it is associated to a non-covariant derivative, which depends on the coordinates. So when you change coordinates you are forced to change the Christoffel tensor as well. See page 34 of his book, after equation 3.1.15. But I know there's confusion about it, and I myself am not totally clear on the situation.

Any insight is helpful...

 

[bcrowell]

It's the collisionless Boltzmann equation:

$$p^\mu \frac{\partial f}{\partial x^\mu} - \Gamma_{\alpha \beta}^\gamma p^\alpha p^\beta \frac{\partial f}{\partial p^\gamma} = 0$$

The first term is manifestly covariant, and so should be the second one.

No, the first term is not manifestly covariant, because it's a partial derivative, not a covariant derivative.

As for the Christoffel symbol, Wald states it is a tensor. The reason why it does not transform like one might suspect is that it is associated to a non-covariant derivative, which depends on the coordinates. So when you change coordinates you are forced to change the Christoffel tensor as well. See page 34 of his book, after equation 3.1.15. But I know there's confusion about it, and I myself am not totally clear on the situation.

Wald calls it a tensor, but says that it doesn't transform as a tensor. Most people only define a thing to be a tensor if it transforms as a tensor.

 

[DrFaustus]

The first term is manifestly covariant because f is a scalar and for scalar quantities the covariant derivative coincides with the usual partial derivative...

 

[George Jones]

The first term is manifestly covariant, and so should be the second one.

I don't quote see the covariance of the second terrm right now, and I probably won't get any chance to think about this tomorrow. I can see that the second term comes from the geodesic condition.

Wald calls it a tensor, but says that it doesn't transform as a tensor. Most people only define a thing to be a tensor if it transforms as a tensor.

Some people make a distinction between a tensor and a tensor field.

 

[bcrowell]

The first term is manifestly covariant because f is a scalar and for scalar quantities the covariant derivative coincides with the usual partial derivative...

Ah, sorry -- you're right and I was wrong.

 

[arkajad]

This is an equation on the tangent bundle. Notice that you have derivation with respect to the coordinates of the tangent vector $$p^\alpha$$ and not with respect to $$x^\alpha$$. The transformation rule involves then second derivatives.

 

[DrFaustus]

george.jones -> Good observation, the second term comes from the geodesic condition indeed.

arkajad -> That's precisely what I'd like to understand better. I know it's a derivative wrt $$p^\alpha$$, which is why I think the full equation should be covariant. Could you give me some more details or a reference where to look up this stuff? What is the transformation rule you mention?

 

[arkajad]

I can' find it right away in a book, but the idea is simple. You have

$$p^\alpha=\frac{\partial x^\alpha}{\partial x^{\alpha'}}p^{\alpha'}$$

therefore

$$\frac{\partial f(x,p)}{\partial x^{\alpha'}} =\frac{\partial f}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial x^{\alpha'}} +\frac{\partial f}{\partial p^\alpha}\frac{\partial p^\alpha}{\partial x^{\alpha'}}$$

Now, the last term gives you

$$\frac{\partial p^\alpha}{\partial x^{\alpha'}}=\frac{\partial^2 x^\alpha}{\partial x^{\alpha'}\partial x^{\beta'}}\,p^{\beta'}$$

This should cancel out with what comes from the second term - though one will need to play with derivatives of the inverse matrix - a little calculational trick is needed (if you want, I can explain it).